If $d_1 =1, d_2, \cdots, d_k = n$ are positive divisors of natural $n$, show $(d_1d_2 \cdots d_k)^2 = n^k$.

Question

If $d_1 =1, d_2, \cdots, d_k = n$ are the positive divisors of the natural number $n$, show that $(d_ld_2 \cdots d_k)^2 = n^k$.

Seems a confusing question with no place to start with, please help. I see no approach except induction to work. Apart from induction, if some way to bring up a proof, then would be quite fine.

Answer 1

In this case the induction is just multiplication, so you can "unroll" it as follows:

Let the prime factorization of $n$ be $n=p_1^{a_1}...p_r^{a_r}$. Then the divisors are $D= \{p_1^{b_1}...p_r^{b_r}:0\leq b_i\leq a_i\}$ with $k=|D|=(a_1+1)...(a_r+1)$.

Multiplying them all together with the fact that the "expected" exponent of $p_i$ is $a_i/2$ (*),

$$(\prod_{d\in D} d)^2=\prod_{d\in D} d^2=(p_1^{a_1/2}...p_r^{a_r/2})^{2|D|}=(p_1^{a_1}...p_r^{a_r})^k=n^k$$

(*) You can think of an element in $D$ as a tuple $(b_1,...b_r)$ of exponents, then $D$ is a cartesian product $\{0,1,...a_1\}\times ... \times \{0,1,...a_r\}$. Multiplying elements of D is like adding their exponents, and it's straightforward to show the average tuple is $(a_1/2,...a_r/2)$.

Answer 2

Let n =1.n =a.b =c.d =e.f =....... = x.y be all possible ways of writing n where 1,n,a,b,c,d, .....,q,r,.......,x,y represent k factors (even number) in all. (this is true even if n may be a perfect square where say n=q.r and q=r)

The k factors are organized into k/2 pairs of divisors such that the product of every pair is equal to n.

so the product of all divisors will be equal to (n)^ (k/2)

so the square of the product of all divisors will be equal to (n)^k